[tex]2. a)\\
\frac{2}{3}\ \textgreater \ \frac{n}{3}\\
Numitorii\ sunt\ la fel,\ deci\ se\ ignora\\
2\ \textgreater \ n\\
n\ \textless \ 2,n\neq0\\
n=1\\\\
b)\\
\frac{5}{4}\ \textless \ \frac{5}{n}\\
Numaratorii\ sunt\ la\ fel,\ deci\ se\ ignora\ si\ se\ schimba\\\
semnul\\
4\ \textgreater \ n\\
n\ \textless \ 4,n\neq0\\
n\in\{1;2;3\}[/tex]
[tex]3.a)\\ Notam\ numarul\ penarelor\ cu\ p \\Notam\ numarul\ \ \ \ cartilor\ cu\ c\\\\ 2p+5c=95\\ 2p+3c=65\\ ------\ Scadem\ cele\ doua\ ecuatii\\ 2p+5c-(2p+3c)=95-65\\ 2p+5c-2p-3c=30 \\5c-3c=30\\ 2c=30\\ c=15\ (o\ carte\ costa\ 15\ lei) [/tex]
[tex]\\\\b)\\
Inlocuim\ in\ prima\ ecuatie\\
2p+5c=95\\
2p+5\cdot15=95\\
2p+75=95\\
2p=95-75\\
2p=20\ (doua\ penare\ costa\ 20\ de\ lei)\\
1p=10\\\\
c) \\4p+7c=\\
=4*10+7*15\\
=40+105\\
=145\\\\
Sper\ ca\ te-am\ ajutat! :)[/tex]
